What is the differnece in shades
1 answer:
You might be interested in
Answer:
b) k Δx - W cos θ - μ mg cos θ = m a , c) θ = 86.6º, d) Δx = 1.18 m
Explanation:
a) In the attachment we can see a diagram of the forces in this problem and the coordinate axes for its decomposition.
F is the force applied by the spring, while it is compressed, this force disappears when the block leaves the spring
b) Let's apply Newton's second law for when the spring is compressed
let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Y axis
N - W_y = 0
N = W_y
N = W cos θ
X axis
F -Wₓ -fr = ma
the force applied by the spring is given by hooke's law
F = k Δx
friction force has the expression
fr = μ N
fr = μ W cos θ
we substitute
k Δx - W cos θ - μ mg cos θ = m a ( 1)
c) If the plane has no friction, what is the angle so that Δx = 0.1m
We write the equation 1, with fr = 0 and since the system is still a = 0
k Δx - W cos θ -0 = 0
cos θ =
cos θ =
cos θ = 0.0598
θ = cos⁻¹ 0.0598
θ = 86.6º
d) In this part they give the angle θ = 45º and there is no friction, they ask the compression
the acceleration is zero, we substitute in 1
k Δx - W cos θ - 0 = 0
Δx =
Δx =
Δx = 1.18 m
Answer:
Explanation:
First of all we shall calculate electric field near charge 2Q .
electric field due to charge Q = K x Q / (5 x 10⁻² )²
E₁ = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis
E₁ = KQ x 10⁴ i / 25
Similarly electric field due to charge 3Q near 2Q
= 3KQ x 10⁴ i / 25 . It is acting along y-axis
E₂ = 3KQ x 10⁴ j / 25
Similarly electric field due to charge 4Q near 2Q
= 4KQ x 10⁴ j / (25 x 2 )
= 2 KQ x 10⁴ / 25 . It is acting acting along north east direction
unit vector in north east direction = ( i + j )/ √2
So E₃ can be represented by
E₃ = 2 KQ x 10⁴ ( i + j ) / 25 x √2
Total field = KQ x 10⁴ i / 25 + 3KQ x 10⁴ j / 25 + 2 KQ x 10⁴ ( i + j ) / ( 25 x √2 )
= KQ x 10⁴ [ i + 3 j + √2 i + √2 j ) / 25
= 400 KQ ( 2.414 i + 4.414 j ) N / C
Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j ) x 2Q N
= 800 KQ² ( 2.414 i + 4.414 j ) N
= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N
= 108.63 ( i + 2 j ) N .
Magnitude of this force
= 108.63 x √5
= 243 N approx .
